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Unit Digit of (3^(4x+2) +7)/10 using cyclicity

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Unit Digit of (3^(4x+2) +7)/10 using cyclicity [#permalink] New post 18 Jul 2012, 02:12
Hi,

Request your help with the following question.

Unit Digit of (3^(4x+2) +7)/10 using cyclicity

There is no official answer available,all I know is that the remainder arrived at is 6.

Thanks..
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Re: Unit Digit of (3^(4x+2) +7)/10 using cyclicity [#permalink] New post 18 Jul 2012, 07:34
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Re: Unit Digit of (3^(4x+2) +7)/10 using cyclicity [#permalink] New post 19 Jul 2012, 00:56
{ 3^(4x + 2) + 7 } / 10

3^4x will always have 1 in the units digit as the unit's digit of pwoers of 3 follow the following sequence
3 9 7 1 3 9 7 1...
So units digit of 3^4x will always be 1
=> units digit of 3^(4x + 2) = 9
So, units digit of 3^(4x + 2) + 7 = 9 + 7 = 6
When you divide any number by 10 then reminder will be the unit's digit of that number
S0, unit's digit of (3^(4x + 2) + 7)/10 will be 6
Hope it helps!
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Re: Unit Digit of (3^(4x+2) +7)/10 using cyclicity [#permalink] New post 29 Jul 2012, 21:43
For 3^x the units digit repeats itself after every 4 numbers producing the pattern {3, 9, 7, 1....3, 9, 7, 1....}. The most confusing part of this equation is determining the units digit that results from 3^(4x+2). Based on the above pattern, you know that the 4x restarts the pattern 1:3, 2:9; 3:7; and 4:1 regardless of the value of x. Therefore adding two will result in the second term (9 in our case). Now we know that the units digit of 3^(4x+2) will be 9, and 9+7 =16. Only the 6 is considered in the last digit so we have ***6/10. The remainder is 6 because of the way our numbering system works (remember it's base 10). e.g. 16 is really 1(10^1) + 6(10^0).
Re: Unit Digit of (3^(4x+2) +7)/10 using cyclicity   [#permalink] 29 Jul 2012, 21:43
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